Most proofs of the characteristic polynomial of the companion matrix–an important specific case–proceed by induction, and start with a 2\times2 matrix.  It strikes me that an inductive proof has more force (or at least makes more sense) if a larger matrix is used.  In this case we will use a “large” (numerical analysts will laugh at this characterisation) 10\times10 matrix.

Let us begin by making a notation change. Consider the general polynomial


For this to be monic (one of the requirements for the polynomial in question) we should divide by the last coefficient, thus


Our object is thus to prove that this (or a variation of this, as we will see) is the characteristic polynomial of


The characteristic polynomial of this is the determinant of the following:


(For another application of the characteristic polynomial and the companion matrix, click here.)

To find the determinant, we expand along the first row. But then we discover that only two minors that matter: the one in the upper left corner and the one in the upper right. Breaking this up into minors and cofactors yields the following:


The second matrix, however, is an upper triangular matrix with ones for all of its diagonal entries. Its determinant, therefore, is unity. Also rewriting the coefficient of the second term, we have




Repeating this process for the next set of minors and cofactors yields


Note carefully the inclusion of -\lambda in the second term. We can also write this as


Repeating this process until the end, it is easy to see that


or more generally


where n is the degree of the polynomial (and the size of the companion matrix.) If we drop the terms we used to make the polynomial monic, we have at last